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1.在重要不等式|a+b|≤|a|+|b|中,当且仅当a≥0,b≥0或a≤0,b≤0时等号成立,即|a+b|=|a|+|b|的充要条件是ab≥0。因此|a+b|<|a|+|b|的充要条件是ab<0。同样,等式|a_1+a_2+…+a_n|=|a_1|+|a_2|+…+|a_n|成立的充要条件是a_1,a_2,…,a_n有相同符号。这一简单事实,在数学中有着重要的应用。 1)在解方程中的应用解方程|lg(2x-3)+lg(4-x~2)|==|lg(2x-3)|+|lg(4-x~2)|。解:根据|a+b|=|a|+|b|的充要条件是ab≥0,所以原方程等价于不等式 lg(2x-3)lg(4-x~2)≥0。解这个不等式: lg(2x-3)lg(4-x~2)≥0 lg(2x-3)lg(4-x~2)≥0 2x-3>0 4-x~2>0
1. In the important inequality |a+b|≤|a|+|b|, if and only if a ≥ 0, b ≥ 0 or a ≤ 0, b ≤ 0, the equal sign is established, ie |a+b| The necessary and sufficient condition for =|a|+|b| is ab≥0. Therefore, the necessary and sufficient condition for |a+b|<|a|+|b| is ab<0. Similarly, the necessary and sufficient condition that the equation |a_1+a_2+...+a_n|=|a_1|+|a_2|+...+|a_n| is established is that a_1, a_2,..., A_n have the same symbol. This simple fact has important applications in mathematics. 1) Apply the solution equation in the solution equation |lg(2x-3)+lg(4-x~2)|==|lg(2x-3)|+|lg(4-x~2)|. Solution: According to |a+b|=|a|+|b|, the necessary and sufficient condition is ab≥0, so the original equation is equivalent to the inequality lg(2x-3)lg(4-x~2)≥0. Solve this inequality: lg(2x-3)lg(4-x~2)≥0 lg(2x-3)lg(4-x~2)≥0 2x-3>0 4-x~2>0